What are the solutions to the equation 4y^2=-12y ?

The solutions to the equation 4y^2=-12y are y=0,y=-3

Find the zeros of 4y^2=-12y

The zeros of 4y^2=-12y are y=0,y=-3

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4 y^{2} = - 12 y

y = 0, y = - 3

Solution steps

4 y^{2} = - 12 y

Move

12 y

to the left side

4 y^{2} + 12 y = 0

Factor

4 y^{2} + 12 y : 4 y (y + 3)

4 y (y + 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y = 0 or y + 3 = 0

Solve

y + 3 = 0 : y = - 3

The solutions to the quadratic equation are:

y = 0, y = - 3

\frac{2 ( \frac{8}{15} )}{1 - ( \frac{8}{15} ) ^{2}}

2 x + 4 = 5 x - 2

\frac{2}{3} \times \frac{2}{3}

4 x^{2} - y^{6}

r^{2} - 6 r + 7 = 0

What are the solutions to the equation 4y^2=-12y ?
The solutions to the equation 4y^2=-12y are y=0,y=-3
Find the zeros of 4y^2=-12y
The zeros of 4y^2=-12y are y=0,y=-3